NCERT Solutions for Class 10 Maths Chapter 4: Quadratic Equations || CBSE Class 10 Maths Chapter 4 Quadratic Equations solutions Ex 4.3 Math Solution

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NCERT Solutions for Class 10 Maths Chapter 4: Quadratic Equations || CBSE Class 10 Maths Chapter 4 Quadratic Equations solutions Ex 4.3 Math Solution
Exercise 4.3

NCERT Solutions for Class 10 Maths Chapter 4: Quadratic Equations || CBSE Class 10 Maths Chapter 4 Quadratic Equations solutions Ex 4.3 Math Solution

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1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:

(i) $ 2 x^{2}-7 x+3=0 $

Answer

$ 2 x^{2}-7 x+3=0 $
Dividing by coefficient of $ x^{2} $, we get
$ x^{2}-\frac{7}{2} x+\frac{3}{2}=0 $
Adding and subtracting the square of $ \left(\frac{b}{2}\right)=\frac{7}{4} $, we get
$ x^{2}-2 \times \frac{7}{4} \times x+\left(\frac{7}{2}\right)^{2}-\left(\frac{7}{2}\right)^{2}+\frac{3}{2}=0 $
Use the formula $ (a-b)^{2}=a^{2}+b^{2}-2 a b $ to get,
$\left(x-\frac{7}{4}\right)^{2}+\frac{3}{2}-\frac{49}{16}=0$
$\left(x-\frac{7}{4}\right)^{2}+\frac{24-49}{16}=0$
$\left(x-\frac{7}{4}\right)^{2}-\frac{25}{16}=0$
$ \left(x-\frac{7}{4}\right)^{2}-\frac{25}{16}=\left(\frac{5}{4}\right)^{2} $
So,
$ \left(x-\frac{7}{4}\right)= \pm \frac{5}{4} $
When $ x-\frac{7}{4}=-\frac{5}{4} $
$ x=-\frac{5}{4}+\frac{7}{4}=\frac{12}{4}=3 $
Hence the value of $ x $ is 3 and $ \frac{1}{2} $.

(ii) $ 2 x^{2}+x-4=0 $

Answer

$ 2 x^{2}+x-4=0 $
Dividing by coefficient of $ x^{2} $
$ x^{2}+\frac{x}{2}-2=0 $
Adding and subtracting the square of $ \frac{b}{2}=\frac{1}{4} $, we get
$ x^{2}+2(x) \times \frac{1}{4}+\left(\frac{1}{4}\right)^{2}-2=0 $
Use the formula $ (a+b)^{2}=a^{2}+b^{2}+2 a b $ to get,
$ \Rightarrow\left(x+\frac{1}{4}\right)^{2}-\left(\frac{1}{16}+2\right)=0 $
$ \Rightarrow\left(x+\frac{1}{4}\right)^{2}=\frac{33}{16} $
$ \Rightarrow x+\frac{1}{4}= \pm \frac{\sqrt{33}}{16} $
$ \Rightarrow x=\frac{-1}{4} \pm \frac{\sqrt{33}}{4} $
$ \Rightarrow x=\frac{-1 \pm \sqrt{33}}{4} $
$ \Rightarrow x=\frac{-1+\sqrt{33}}{4} $
Or $ x=\frac{-1-\sqrt{33}}{4} $

(iii) $ 4 x^{2}+4 \sqrt{3} x+3=0 $

Answer

$ 4 x^{2}+4 \sqrt{3} x+3=0 $
Divide the whole equation by 4 to get, $ \frac{4 x^{2}+4 \sqrt{3} x+3}{4}=\frac{0}{4} $
Add and subtract the square of half of the coefficient of $ x $,
$\Rightarrow x=x^{2}+\sqrt{3} x+\frac{3}{4}=0$
$\Rightarrow x^{2}+\sqrt{3} x+\frac{3}{4}+\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2}=0$
Use the formula $ (a+b)^{2}=a^{2}+b^{2}+2 a b $ to get,
$\Rightarrow\left(x+\frac{\sqrt{3}}{2}\right)^{2}+\frac{3}{4}-\left(\frac{\sqrt{3}}{2}\right)^{2}=0$
$\Rightarrow\left(x+\frac{\sqrt{3}}{2}\right)^{2}+\frac{3}{4}-\frac{3}{4}=0$
$ \Rightarrow\left(x+\frac{\sqrt{3}}{2}\right)^{2}= \pm 0 $
$ \Rightarrow x+\frac{\sqrt{3}}{2}=0 $ and $ x+\frac{\sqrt{3}}{2}=-0 $
$ \Rightarrow x=0-\frac{\sqrt{3}}{2} $ and $ x=0-\frac{\sqrt{3}}{2} $
$ \Rightarrow x=-\frac{\sqrt{3}}{2} $ and $ x=-\frac{\sqrt{3}}{2} $
$ \therefore $ Roots of equation are $ =x=-\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2} $

(iv) $ 2 x^{2}+x+4=0 $

Answer

$ 2 x^{2}+x+4=0 $
Dividing by coefficient of $ x^{2} $ we get,
$ x^{2}+\frac{x}{2}-2=0 $
Adding and subtracting the square of $ \frac{b}{2}=\frac{1}{4} $, we get
$ \Rightarrow x^{2}+\frac{x}{2}+2+\left(\frac{1}{4}\right)^{2}-\left(\frac{1}{4}\right)^{2}=0 $
$ x^{2}+\frac{x}{2}+2=0 $
Adding and subtracting the square of coefficient of $ x $ we get,
$ \Rightarrow x^{2}+\frac{x}{2}+2+\left(\frac{1}{4}\right)^{2}-\left(\frac{1}{4}\right)^{2}=0 $
Use the formula $ (a+b)^{2}=a^{2}+b^{2}+2 a b $ to get,
$ x^{2}+\frac{x}{2}+2=0$
$\Rightarrow \left(x+\frac{1}{4}\right)^{2}+2-\left(\frac{1}{4}\right)^{2}=0$
$\Rightarrow \left(x+\frac{1}{4}\right)^{2}+2-\frac{1}{16}=0$
$\Rightarrow \left(x+\frac{1}{4}\right)^{2}+\frac{32-1}{16}=0$
$ \Longrightarrow\left(x+\frac{1}{4}\right)^{2}+\frac{31}{16}=0$
$\Rightarrow\left(x+\frac{1}{4}\right)^{2}=-\frac{31}{16}$
Square of a number cannot be negative, Hence roots of the
given equation do not exist.

NCERT Solutions for Class 10 Maths Chapter 4: Quadratic Equations || CBSE Class 10 Maths Chapter 4 Quadratic Equations solutions Ex 4.3 Math Solution

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4. The sum of the reciprocals of Rehman's ages, (in years) 3 years ago and 5 years from now is $ \frac{1}{3} $ Find his present age.

Answer

Let the present age of Rehman be $ x $ years.
Three years ago, his age was $ (x-3) $ years.
Five years hence, his age will be $ (x+5) $ years.
It is given that the sum of the reciprocals of Rehman's ages 3
years ago and 5 years from now is $ \frac{1}{3} $.
$\therefore \frac{1}{x-3}+\frac{1}{x+5}=\frac{1}{3}$
$\Rightarrow \quad \frac{x+5+x-3}{(x-3)(x+5)}=\frac{1}{3}$
$\Rightarrow \quad \frac{2 x+2}{(x-3)(x+5)}=\frac{1}{3}$
$\Rightarrow 3(2 x+2)=(x-3)(x+5)$
$\Rightarrow 6 x+6=x^{2}+2 x-15$
$\Rightarrow x^{2}-4 x-21=0$
$\Rightarrow x^{2}-7 x+3 x-21=0$
$\Rightarrow x(x-7)+3(x-7)=0$
$\Rightarrow x=7,-3$
However, age cannot be negative.
Therefore, Rehman's present age is 7 years.

5. In a class test, the sum of Shefali's marks in Mathematics and English is 30 . Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

Answer

Let the marks obtained in Mathematics by Shefali be ' $ a $ '.
Given, sum of the marks obtained by Shefali in Mathematics and English is 30 .
Marks obtained in english $ =30-\mathrm{a} $
Also, she got 2 marks more in Mathematics and 3 marks less
in English, the product of her marks would have been 210 .
$ \Rightarrow(a+2)(30-a-3)=210$
$\Rightarrow(a+2)(27-a)=210$
$\Rightarrow 27 a-a^{2}+54-2 a=210$
$\Rightarrow-a^{2}+25 a+54=210$
$\Rightarrow-a^{2}+25 a+54-210=0$
$\Rightarrow-a^{2}+25 a-156=0$
$\Rightarrow a^{2}-25 a+156=0$
In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.
$\Rightarrow a^{2}-13 a-12 a+156=0$
$\Rightarrow a(a-13)-12(a-13)=0$
$\Rightarrow(a-12)(a-13)=0$
$\Rightarrow a=12 \text { or } 13$
If marks in mathematics is 12
marks in english is $ 30-\mathrm{a}=30-12=18 $
If marks in mathematics is 13 marks in english is $ 30-\mathrm{a}=30-13=17 $
Hence
Marks in Mathematics $ =12 $, Marks in English $ =18 $
Or
Marks in Mathematics $ =13 $, Marks in English $ =17 $

6. The diagonal of a rectangular field is $60$ metres more than the shorter side. If the longer side is $30$ metres more than the shorter side, find the sides of the field.

Answer

Let the shorter side of the rectangle be $ x \mathrm{~m} $.
Then, larger side of the rectangle $ =(x+30) \mathrm{m} $
we know,
By pythagoras theorem,
$ (\text { Hypotenuse })^{2}=(\text { Base })^{2}+(\text { Perpendicular })^{2} $
Diagonal of a rectangle is $ \sqrt{ }\left[(\text { length })^{2}+(\text { breadth })^{2}\right] $
Diagonal of the rectangle $ =\sqrt{x^{2}+(x+30)^{2}} $
It is given that the diagonal of the rectangle is 60 m more than the shorter side.
$ \sqrt{x^{2}+(x+30)^{2}}=x+60 $
Squaring both sides, we get,
$\Rightarrow x^{2}+(x+30)^{2}=(x+60)^{2}$
$\Rightarrow x^{2}+x^{2}+900+60 x=x^{2}+3600+120 x$
$\Rightarrow x^{2}-60 x-2700=0$
Now for solving this quadratic equation, we need to factorize
60 in such a way that the product is 2700 and the difference is 60
$ \Rightarrow x(x-90)+30(x-90)=0 $
$\Rightarrow(x-90)(x+30)=0$
$\Rightarrow x=90,-30$
However, side cannot be negative. Therefore, the length of the shorter side will be 90 m .
Hence, length of the larger side will be $ (\mathbf{9 0}+\mathbf{3 0}) \mathbf{m}=\mathbf{1 2 0} \mathrm{m} $

NCERT Solutions for Class 10 Maths Chapter 4: Quadratic Equations || CBSE Class 10 Maths Chapter 4 Quadratic Equations solutions Ex 4.3 Math Solution

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7. The difference of squares of two numbers is $ 180$. The square of the smaller number is $8$ times the larger number.
Find the two numbers.

Answer

Assumption: Let the larger and smaller number be x and y respectively.
Given:
According to the given question difference of squares of two numbers is 180 and the square of smaller number is 8 times square of the larger number.
So,
$ \Rightarrow x^{2}-\mathrm{y}^{2}=180 \quad \text { eq(i) }$
$\Rightarrow \mathrm{y}^{2}=8 x \quad \text { eq(ii) }$
putting value of eq(ii) in eq(i)
$\Rightarrow x^{2}-8 x=180$
$\Rightarrow x^{2}-8 x-180=0$
For factorizing this quadratic equation, the product of numbers should be 180 and their difference should be 8
$\Rightarrow x^{2}-18 x+10 x-180=0$
$\Rightarrow x(x-18)+10(x-18)=0$
$\Rightarrow(x-18)(x+10)=0$
$\Rightarrow x=18,-10$
The larger number cannot be negative as it makes the square of smaller number negative, which is not possible.
Therefore the larger number will be 18 only.
$ x=18$
$\therefore \mathrm{y}^{2}=8 x$
$=8 \times 18=144$
$=y= \pm \sqrt{144}= \pm 12$
$\therefore \text { smaller number }= \pm 12$
Therefore, the numbers are 18 and 12 or 18 and -12.

8. A train travels $ 360$ km at a uniform speed. If the speed had been $ 5 \mathrm{~km} / \mathrm{h} $ more, it would have taken $ 1$ hour less for the same journey. Find the speed of the train.

Answer

Let the speed of the train be $ x \mathrm{~km} / \mathrm{hr} $.
Time taken to cover $ 360 \mathrm{~km}=\frac{360}{x} \mathrm{hr} $, As time=$\frac{\text { distance }}{\text { speed }} $
Now, given that if the speed would be $ 5 \mathrm{~km} / \mathrm{hr} $ more, the same distance would be covered in 1 hour less, i.e. if speed $ =x+5 $, and time= $ \left(\frac{360}{x}-1\right) $
then, using distance $ = $ speed x time, we have
$ (x+5)\left(\frac{360}{x}-1\right)=360 $
Now we can form the quadratic equation from this equation
$360-x+\frac{1800}{x}-5=360$
$\frac{360 x-x^{2}+1800-5 x}{x}=360$
Now, cross multiplying we get $ \Rightarrow 360 x-x^{2}+1800-5 x=360 x $
$ \Rightarrow x^{2}+5 x-1800=0 $
Now we have to factorize in such a way that the product of
the two numbers is $1800$ and the difference is $ 5 \Rightarrow x^{2}+45 x-40 x-1800=0 $
$\Rightarrow x(x+45)-40(x+45)=0$
$\Rightarrow(x+45)(x-40)=0$
$\Rightarrow x=-45,40$
since, the speed of train can't be negative, so, speed will be $ 40 \mathrm{~km} / \mathrm{hour} $.

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9. Two water taps together can fill a tank in hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Answer

Let the time taken by the smaller pipe to fill the tank be x hr . Time taken by the larger pipe $ =(x-10) \mathrm{hr} $
Part of the tank filled by a smaller pipe in 1 hour $ =\frac{1}{x} $
Part of the tank filled by the larger pipe in 1 hour $ =\frac{1}{x-10} $
It is given that the tank can be filled in $ 9 \frac{3}{8}=\frac{75}{8} $ hours by both the pipes together.
So $ 75 / 8 $ hours, multiplied by the sum of parts filled with both pipes in one hour equal to complete work i.e 1.
$ \frac{75}{8}\left(\frac{1}{x}+\frac{1}{x-10}\right)=1$
$\Rightarrow \frac{1}{x}+\frac{1}{x-10}=\frac{8}{75}$
$\Rightarrow \frac{x-10+x}{x(x-10)}=\frac{8}{75}$
$\Rightarrow \frac{2 x-10}{x(x-10)}=\frac{8}{75}$
$\Rightarrow 75(2 x-10)=8 x^{2}-80 x$
$\Rightarrow 150 x-750=8 x^{2}-80 x$
$\Rightarrow 8 x^{2}-230 x+750=0$
Now for factorizing the above quadratic equation, two
numbers are to be found such that their product is equal to $ 750 \times 8 $ and their sum is equal to 230
$ \Rightarrow 8 x^{2}-200 x-30 x+750=0$
$\Rightarrow 8 x(x-25)-30(x-25)=0$
$\Rightarrow(x-25)(8 x-30)=0$
$\Rightarrow x=25, \frac{30}{8}$
Time taken by the smaller pipe cannot be $ \frac{30}{8}=3.75 $ hours.
As in this case, the time taken by the larger pipe will be negative, which is logically not possible.
Therefore, time taken individually by the smaller pipe and the larger pipe will be 25 and $ 25-10=15 $ hours respectively.

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10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is $ 11 \mathrm{~km} / \mathrm{h} $ more than that of the passenger train, find the average speed of the two trains.

Answer

Let the average speed of passenger train be $ x \mathrm{~km} / \mathrm{h} $.
Average speed of express train $ =(x+11) \mathrm{km} / \mathrm{h} $
It is given that the time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance.
As,
time $ =\frac{\text { distance }}{\text { speed }} $
Time taken by passenger train $ =\frac{132}{x} $
Time taken by express train $ =\frac{132}{x+11} $
Now Time taken by passenger train - Time taken by express train $ =1 $
$\therefore \frac{132}{x}-\frac{132}{x+11}=1$
$=132\left[\frac{x+11-x}{x(x+11)}\right]=1$
$\Rightarrow \frac{132 \times 11}{x(x+11)}=1$
$\Rightarrow 132 \times 11=x(x+11)$
$\Rightarrow x^{2}+11 x-1452=0$
Now we have to factorize this equation such that the product
of numbers is 1452 and their difference is 11
$\Rightarrow x^{2}+44 x-33 x-1452=0$
$\Rightarrow x(x+44)-33(x+44)=0$
$\Rightarrow(x+44)(x-33)=0$
$\Rightarrow x=-44,33$
Speed cannot be negative.
Therefore, the speed of the passenger train will be $ 33 \mathrm{~km} / \mathrm{h} $ and thus, the speed of the express train will be $ 33+11=44 \mathrm{~km} / \mathrm{h} $.

11. Sum of the areas of two squares is $ 468 \mathrm{~m}^{2} $. If the difference of their perimeters is 24 m , find the sides of the two squares.

Answer

Let the sides of the two squares be $ x \mathrm{~m} $ and $ y \mathrm{~m} $.
Therefore, their perimeter will be $ 4 x $ and $ 4 y $ respectively and their areas will be $ x^{2} $ and $ y^{2} $ respectively.
It is given that $ 4 x-4 y=24 $ [Difference of perimeter] or $ x-y=6 $
$ x=y+6 $
Also, $ x^{2}+\mathrm{y}^{2}=468 $ [sum of squares is 468]
$ (6+y)^{2}+y^{2}=468 $
$36+y^{2}+12 y+y^{2}=468$
$2 y^{2}+12 y-432=0$
$y^{2}+6 y-216=0$
$y^{2}+18 y-12 y-216=0$
$y(y+18)(y-12)=0$
$y=-18 \text { or } 12$
However, side of a square cannot be negative.
Hence, the sides of the squares are 12m and $ (12+6) \mathrm{m}=18 \mathrm{~m} $

NCERT Solutions for Class 10 Maths Chapter 4: Quadratic Equations || CBSE Class 10 Maths Chapter 4 Quadratic Equations solutions Ex 4.3 Math Solution

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NCERT Solutions for Class 10 Maths Chapter 4: Quadratic Equations || CBSE Class 10 Maths Chapter 4 Quadratic Equations solutions Ex 4.3 Math Solution

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